package com.fishercoder.solutions;

/**
 * 161. One Edit Distance
 *
 * Given two strings s and t, determine if they are both one edit distance apart.
 *
 * Note:
 *
 * There are 3 possiblities to satisify one edit distance apart:
 *
 *     Insert a character into s to get t
 *     Delete a character from s to get t
 *     Replace a character of s to get t
 *
 * Example 1:
 * Input: s = "ab", t = "acb"
 * Output: true
 * Explanation: We can insert 'c' into s to get t.
 *
 * Example 2:
 * Input: s = "cab", t = "ad"
 * Output: false
 * Explanation: We cannot get t from s by only one step.
 *
 * Example 3:
 * Input: s = "1203", t = "1213"
 * Output: true
 * Explanation: We can replace '0' with '1' to get t.
 */
public class _161 {
  public static class Solution1 {
    public boolean isOneEditDistance(String s, String t) {
      char[] schar = s.toCharArray();
      char[] tchar = t.toCharArray();

      if (Math.abs(s.length() - t.length()) == 1) {
        char[] longer = (s.length() > t.length()) ? schar : tchar;
        char[] shorter = (longer == schar) ? tchar : schar;

        int diffCnt = 0;
        int i = 0;
        int j = 0;
        for (; i < shorter.length && j < longer.length; ) {
          if (longer[j] != shorter[i]) {
            diffCnt++;
            j++;
          } else {
            i++;
            j++;
          }
        }
        return diffCnt == 1
            || diffCnt
            == 0;//it could be the last char of the longer is the different one, in that case, diffCnt remains to be zero
      } else if (s.length() == t.length()) {
        int diffCnt = 0;
        for (int i = 0; i < s.length(); i++) {
          if (schar[i] != tchar[i]) {
            diffCnt++;
          }
          if (diffCnt > 1) {
            return false;
          }
        }
        return diffCnt == 1;
      }
      return false;
    }
  }
}
